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Top Forums UNIX for Dummies Questions & Answers blank space in regex pattern using sed Post 302402048 by glev2005 on Monday 8th of March 2010 07:07:06 PM
Old 03-08-2010
blank space in regex pattern using sed

why does sed 's/.* //' show the last word in a line

and

sed 's/ .*//' show the first word in a line? How is that blank space before or after the ".*" being interpreted in the regex?

i would think the first example would delete the first word and the next example would delete the second word because you have .* to match any number of characters and then a space substituted with nothing, should'nt that remove the first matched word on the line that is terminated with whitespace?

---------- Post updated at 07:07 PM ---------- Previous update was at 06:49 PM ----------

I think i get it... if you do /.* // that will grab the first thing on the line if it has no space before it, and replace it will whitespace, and then print the rest of the line,
but if you do / .*// it will ignore what is before the first space, and begin substitution after the first space removing the rest of the line by substituting white space. Correct?
 

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STRSPLIT(3pub)						       C Programmer's Manual						    STRSPLIT(3pub)

NAME
strsplit - split string into words SYNOPSIS
#include <publib.h> int strsplit(char *src, char **words, int maxw, const char *sep); DESCRIPTION
strsplit splits the src string into words separated by one or more of the characters in sep (or by whitespace characters, as specified by isspace(3), if sep is the empty string). Pointers to the words are stored in successive elements in the array pointed to by words. No more than maxw pointers are stored. The input string is modifed by replacing the separator character following a word with ''. However, if there are more than maxw words, only maxw-1 words will be returned, and the maxwth pointer in the array will point to the rest of the string. If maxw is 0, no modification is done. This can be used for counting how many words there are, e.g., so that space for the word pointer table can be allocated dynamically. strsplit splits the src string into words separated by one or more of the characters in sep (or by whitespace characters, as defined by isspace(3), if sep is the empty string). The src string is modified by replacing the separator character after each word with ''. A pointer to each word is stored into successive elements of the array words. If there are more than maxw words, a '' is stored after the first maxw-1 words only, and the words[maxw-1] will contain a pointer to the rest of the string after the word in words[maxw-2]. RETURN VALUE
strsplit returns the total number of words in the input string. EXAMPLE
Assuming that words are separated by white space, to count the number of words on a line, one might say the following. n = strsplit(line, NULL, 0, ""); To print out the fields of a colon-separated list (such as PATH, or a line from /etc/passwd or /etc/group), one might do the following. char *fields[15]; int i, n; n = strsplit(list, fields, 15, ":"); if (n > 15) n = 15; for (i = 0; i < n; ++i) printf("field %d: %s ", i, fields[i]); In real life, one would of course prefer to not restrict the number of fields, so one might either allocated the pointer table dynamically (first counting the number of words using something like the first example), or realize that since it is the original string that is being modified, one can do the following: char *fields[15]; int i, n; do { n = strsplit(list, fields, 15, ":"); if (n > 15) n = 15; for (i = 0; i < n; ++i) printf("field %d: %s ", i, fields[i]); list = field[n-1] + strlen(field[n-1]); } while (n == 15); SEE ALSO
publib(3), strtok(3) AUTHOR
The idea for this function came from C-News source code by Henry Spencer and Geoff Collyer. Their function is very similar, but this implementation is by Lars Wirzenius (lars.wirzenius@helsinki.fi) Publib C Programmer's Manual STRSPLIT(3pub)
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